Can anyone help with this question?

Let f and g be multiplicative functions that are not identically zero and such that f(p^k)=g(p^k) for each prime p and k≥1. Prove that f = g.

Results 1 to 4 of 4

- April 2nd 2007, 10:15 AM #1

- Joined
- Feb 2007
- Posts
- 7

- April 2nd 2007, 10:20 AM #2

- Joined
- Nov 2005
- From
- New York City
- Posts
- 10,616
- Thanks
- 10

We want to show that,

f(n)=g(n) for any positive integer n.

Your statement is not actually true . It is true if n>1.

You need an addition fact that f(1)=g(1).

But anyway, let me show that they agree on n>1.

If n>1 we can prime factorize,

n=p_1^(a_1) * p_2^(a_2) * ... * p_k^(a_k)

Now,

f(n)=f(p_1^(a_1)*p_2^(a_2)*...*p_k^(a_k))=f(p_1^(a _1))*...*f(p_k^(a_k)).

Because,

gcd(p_i,p_j)=1 if i!=j.

The same thing for g(n).

But,

f(p_i^(a_i))=g(p_i^(a_i))

Thus,

f(n)=g(n) for n>1.

- April 2nd 2007, 10:29 AM #3

- Joined
- Feb 2007
- Posts
- 7

- April 2nd 2007, 12:06 PM #4

- Joined
- Nov 2005
- From
- New York City
- Posts
- 10,616
- Thanks
- 10

Because,

f(p_1^(a_1))=g(p_1^(a_1))

f(p_2^(a_2))=g(p_2^(a_2))

...

f(p_k^(a_k))=g(p_k^(a_k))

Then multily them out,

f(p_1^(a_1))*...*f(p_k^(a_k))=g(p_1^(a_1))*...*g(p _k^(a_k))

Now you can put them under the same function because it is weakly multiplicative.

f(p_1^(a_1)*...*p_k^(a_K))=g(p_1^(a_1)*...*p_k^(a_ k))

But then,

f(n)=g(n)